-4
$\begingroup$

I am not very experienced at solving inequalities and I cannot understand the following: I was given the inequality

$$\tag{1} 2xy>x^2-y^2.$$

When I put this into WolframAlpha I get $$ y>\sqrt2 \sqrt{x^2} - x \text{ and } y<-\sqrt2 \sqrt{x^2} - x$$

How do I come to these two inequalities from $(1)$?

$\endgroup$

closed as off-topic by user21820, Dmoreno, Mostafa Ayaz, callculus, Ethan Bolker Feb 21 '18 at 17:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Dmoreno, callculus, Ethan Bolker
If this question can be reworded to fit the rules in the help center, please edit the question.

-3
$\begingroup$

[Edited after typos in OP are corrected.]

When I ask WolframAlpha I get the answer $y>\sqrt2 \sqrt{x^2} - x$ and $y<-\sqrt2 \sqrt{x^2} - x$. How do I find this solution?

WolframAlpha shows you:

enter image description here

To see how the results are found, taking $x$ as a constant and solving the quadratic equation in $y$: $$ y^2+2xy-x^2=0\tag{*} $$ one has $$ \frac{-2x\pm\sqrt{4x^2+4x^2}}{2}=-x\pm\sqrt{2}\sqrt{x^2}=-x\pm\sqrt{2}|x|. $$ One can recognize these two roots in the "Results" above.


Note that the two roots can be written as (regardless $x$ is positive or not) $$ y_1=(-1+\sqrt{2})x,\quad y_2=(-1-\sqrt{2})x. $$

  • When $x\geqslant 0$, $y_1\geqslant y_2$ and the inequality $y^2+2xy-y^2>0$, which is equivalent to $(y-y_1)(y-y_2)>0$, implies that $$ y>y_1\quad \hbox{or}\quad y<y_2. $$

  • When $x<0$, $y_1<y_2$. Now the inequality $y^2+2xy-y^2>0$ implies that $$ y>y_2\quad \hbox{or}\quad y<y_1. $$

Combining these two cases together, one can see that for $x\in{\bf R}$, $$ y<-x-\sqrt{2}|x|\qquad\hbox{or}\quad y>-x+\sqrt{2}|x|. $$

$\endgroup$
  • $\begingroup$ Alright I think I understand this answer. Thank you! I also know that x>0 and y>0, what impact will this have on the answer? The last case is not true I suppose? What about when x>y? I agree about the negative votes, very strange! I wonder why my question was downvoted so much. $\endgroup$ – Emilia314 Sep 30 '17 at 18:16
  • $\begingroup$ @Mathmellow Solving the inequality $2xy>x^2-y^2$ really means finding the set $S=\{(x,y)\in{\bf R}^2\mid 2xy>x^2-y^2\}$. Note that WolframAlpha tells you that $S=\{(x,y)\in{\bf R}^2\mid y<-x-\sqrt{2}|x|\ \hbox{ or}\ y>-x+\sqrt{2}|x|\}$ (and you can see the picture in Will Jay's answer.) If you further assume that $x>0$ and $y>0$, the solution set would become $S\cap Q$ where $Q=\{(x,y)\mid x>0\ \hbox{and }y>0\}$, similarly for other constraints. Does this answer your question? $\endgroup$ – Jack Sep 30 '17 at 18:22
  • 3
    $\begingroup$ Yes, I think so. Will's answer was also very intuitive and helpful. May I type up my answer here as well so other people may comment on it, in case I misunderstood? $\endgroup$ – Emilia314 Sep 30 '17 at 18:47
  • $\begingroup$ @Mathmellow: you could certainly post your own answer. $\endgroup$ – Jack Sep 30 '17 at 18:59
5
$\begingroup$

Hint:

If $y=0$, we have $x^2<0$, which is impossible for real $x$.

So dividing both sides by $y^2$, we get $$ \left(\dfrac xy\right)^2-2\left(\dfrac xy\right)-1<0. $$

On the other hand, for $(z-a)(z-b)<0$ with $a<b$, we can prove that $$ a<z<b. $$

$\endgroup$
  • $\begingroup$ What are $a$ and $b$? $\endgroup$ – Jack Sep 30 '17 at 16:22
  • 1
    $\begingroup$ @Jack The roots of the quadratic Lab decscribed $\endgroup$ – Jyrki Lahtonen Sep 30 '17 at 16:26
  • $\begingroup$ @JyrkiLahtonen: which quadratic equation are you referring to? $\endgroup$ – Jack Sep 30 '17 at 16:27
  • 1
    $\begingroup$ @Jack $$\left(\frac xy\right)^2-2\left(\frac xy\right)-1.$$ $\endgroup$ – Jyrki Lahtonen Sep 30 '17 at 16:29
  • 2
    $\begingroup$ @Jack then shouldn't it be $(\frac{x}{y}-a)(\frac{x}{y}-b)<0$? Exactly! That's Lab's point, I think. He was just describing the set of solutions of a quadratic inequality. That he used $x/y$ as the variable in one and $x$ in the other is inconsequential. The ratio $x/y$ is what shows in this problem, the other was a generic quadratic. $\endgroup$ – Jyrki Lahtonen Sep 30 '17 at 16:48
3
$\begingroup$

You want $y^2 + 2 yx - x^2 > 0.$ The boundaries are $y^2 + 2 yx - x^2 = 0.$ Let the slope of a line be $$ m = \frac {y}{x}, $$ divide $y^2 + 2 yx - x^2 = 0$ by $x^2$ for nonzero $x,$ giving $$ m^2 + 2m - 1 = 0. $$

Put another way, you want $$ \left( y - \left( \sqrt 2 - 1 \right) x \right) \left( y + \left( \sqrt 2 + 1 \right) x \right) > 0 $$ Either both factors are positive or both negative. We can see which of the four quarters in the picture work by noticing that $(0,2)$ works, and $(0,-2)$ works. The borders are the depicted slanted lines. The slanted lines are called the null cone of the (indefinite) quadratic form $y^2 + 2 yx - x^2.$

enter image description here enter image description here

$\endgroup$
1
$\begingroup$

$2xy\gt x^2-y^2; $

$2y/x \gt 1 - (y/x)^2$.

$[y/x +1]^2 \gt 2.$

$(y+x)^2 \gt 2x^2.$

1)$ y+x \ge 0: y\ge -x.$

$y+x \gt \sqrt{2x^2}.$

$y \gt \sqrt{2x^2} -x.$

2) $y+x \lt 0; y \lt -x.$

$-(y+x) \gt \sqrt{2x^2}$.

$y \lt -\sqrt{2x^2} - x$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.